Question

我正在研究家庭树木：

我根据sqldf https://www.r-bloggers.com/exploring-recursive-ctes-with-sqldf/

改编了Bob Horton的例子

我的数据：

POST to http://my-api/class/learn-rest/enrollment

我的结果，后代按“Guillou Arthur”（没有父亲的顶级人物）的等级排序：

      person            father
      Guillou Arthur    NA          
      Cleach Marc       NA          
      Guillou Eric      Guillou Arthur          
      Guillou Jacques   Guillou Arthur          
      Cleach Franck     Cleach Marc         
      Cleach Leo        Cleach Marc         
      Cleach Herbet     Cleach Leo          
      Cleach Adele      Cleach Herbet           
      Guillou Jean      Guillou Eric            
      Guillou Alan      Guillou Eric

您可以使用sqldf：

进行递归查询来构建此表

数据：

  name    parent_name              level
  Guillou Arthur    NA                  1       
  Guillou Eric      Guillou Arthur      2       
  Guillou Jacques   Guillou Arthur      2       
  Guillou Alan      Guillou Eric        3       
  Guillou Jean     Guillou Eric         3

大到长格式转换：

 person <- c("Guillou Arthur",
              "Cleach Marc",
              "Guillou Eric",
              "Guillou Jacques", 
              "Cleach Franck",
              "Cleach Leo",
              "Cleach Herbet",
              "Cleach Adele",
              "Guillou Jean",
              "Guillou Alan" )
 father <- c(NA, NA, "Guillou Arthur" , "Guillou Arthur", "Cleach Marc", "Cleach Marc", "Cleach Leo", "Cleach Herbet", "Guillou Eric", "Guillou Eric")


family <- data.frame(person, father)

递归查询以查找“Guillou Arthur”的后代（没有父亲的顶级人物）：

    library(tidyr)

    long_family <- gather(family, parent, parent_name, -person)

    long_family

我的问题：
如何使用R（而不是sql）直接创建包含所有族树的data.frame对象。每棵树都以一个族长（没有父亲）开始，如“Cleach Marc”。（使用R方法或sqldf方法）

Answer 1

我们构建一个递归函数来获取父行，从那里一切都很容易。

首先，我们使用stringsAsFactors = FALSE定义数据，以便更顺畅地重新格式化。

family <- data.frame(person, father,stringsAsFactors = FALSE)

功能

father_line <- function(x){
dad <- subset(family,person==x)$father
if(is.na(dad)) return(x)
c(x,father_line(dad))
}

father_line ("Guillou Alan")
# [1] "Guillou Alan"   "Guillou Eric"   "Guillou Arthur"

使用它来获取关卡和其他内容

family$father_line <- lapply(family$person,father_line)
family$level       <- lengths(family$father_line)
family$patriarch   <- sapply(family$father_line,tail,1)

#             person         father                                          father_line level      patriarch
# 1   Guillou Arthur           <NA>                                       Guillou Arthur     1 Guillou Arthur
# 2      Cleach Marc           <NA>                                          Cleach Marc     1    Cleach Marc
# 3     Guillou Eric Guillou Arthur                         Guillou Eric, Guillou Arthur     2 Guillou Arthur
# 4  Guillou Jacques Guillou Arthur                      Guillou Jacques, Guillou Arthur     2 Guillou Arthur
# 5    Cleach Franck    Cleach Marc                           Cleach Franck, Cleach Marc     2    Cleach Marc
# 6       Cleach Leo    Cleach Marc                              Cleach Leo, Cleach Marc     2    Cleach Marc
# 7    Cleach Herbet     Cleach Leo               Cleach Herbet, Cleach Leo, Cleach Marc     3    Cleach Marc
# 8     Cleach Adele  Cleach Herbet Cleach Adele, Cleach Herbet, Cleach Leo, Cleach Marc     4    Cleach Marc
# 9     Guillou Jean   Guillou Eric           Guillou Jean, Guillou Eric, Guillou Arthur     3 Guillou Arthur
# 10    Guillou Alan   Guillou Eric           Guillou Alan, Guillou Eric, Guillou Arthur     3 Guillou Arthur

例如，要获得规定的预期输出：

subset(family,patriarch == "Guillou Arthur",select=c(person,father,level))
#             person         father level
# 1   Guillou Arthur           <NA>     1
# 3     Guillou Eric Guillou Arthur     2
# 4  Guillou Jacques Guillou Arthur     2
# 9     Guillou Jean   Guillou Eric     3
# 10    Guillou Alan   Guillou Eric     3

tidyverse方式如下：

library(tidyverse)
family %>%
  mutate(family_line = map(person,father_line),
         level = lengths(family_line),
         patriarch = map(family_line,last)) %>%
  filter(patriarch == "Guillou Arthur") %>%
  select(person,father,level)

#            person         father level
# 1  Guillou Arthur           <NA>     1
# 2    Guillou Eric Guillou Arthur     2
# 3 Guillou Jacques Guillou Arthur     2
# 4    Guillou Jean   Guillou Eric     3
# 5    Guillou Alan   Guillou Eric     3

Answer 2

您可以使用图表工具执行此操作。所以使用igraph，您可以使用ego函数来获取邻居。

快速草图（需要检查！）

library(igraph)

family[] = lapply(family, factor, levels=unique(unlist(family)))

g = graph_from_adjacency_matrix(table(family))

cg = connect.neighborhood(g, order=length(V(g)), mode="out")

cbind( V(cg)$name, 
       sapply(ego(g, mode="out", mindist=1), function(x) replace(names(x), length(names(x))==0, NA)),
       ego_size(cg, mode="out") )[grep("Guillou", V(cg)$name),]

[,1]                   [,2]             [,3]
[1,] "Guillou Arthur"  NA               "1" 
[2,] "Guillou Eric"    "Guillou Arthur" "2" 
[3,] "Guillou Jacques" "Guillou Arthur" "2" 
[4,] "Guillou Jean"    "Guillou Eric"   "3" 
[5,] "Guillou Alan"    "Guillou Eric"   "3"

事实上，你可能不需要创建一个邻居图，可以使用：

cbind( V(g)$name, 
       sapply(ego(g, mode="out", mindist=1), function(x) replace(names(x), length(names(x))==0, NA)),
       ego_size(g, mode="out", order=length(V(g))) )[grep("Cleach", V(g)$name),]

在R中构建Family嵌套树父/子关系

2 个答案: